Divide (4y^2-4y+1)/(2y-1)

$\frac{4 y^{2} - 4 y + 1}{2 y - 1}$

Step 1

Factor using the perfect square rule.

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$\frac{{(2 y)}^{2} - 4 y + 1}{2 y - 1}$

Rewrite $1$ as $1^{2}$ .

$\frac{{(2 y)}^{2} - 4 y + 1^{2}}{2 y - 1}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 y = 2 \cdot (2 y) \cdot 1$

Rewrite the polynomial.

$\frac{{(2 y)}^{2} - 2 \cdot (2 y) \cdot 1 + 1^{2}}{2 y - 1}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 y$ and $b = 1$ .

$\frac{{(2 y - 1)}^{2}}{2 y - 1}$

Step 2

Cancel the common factor of ${(2 y - 1)}^{2}$ and $2 y - 1$ .

Factor $2 y - 1$ out of ${(2 y - 1)}^{2}$ .

$\frac{(2 y - 1) (2 y - 1)}{2 y - 1}$

Cancel the common factors.

Multiply by $1$ .

$\frac{(2 y - 1) (2 y - 1)}{(2 y - 1) \cdot 1}$

Cancel the common factor.

$\frac{(2 y - 1) (2 y - 1)}{(2 y - 1) \cdot 1}$

Rewrite the expression.

$\frac{2 y - 1}{1}$

Divide $2 y - 1$ by $1$ .

$2 y - 1$